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Question-212557

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Question Number 212557 by tri26112004 last updated on 17/Oct/24
Answered by Ghisom last updated on 17/Oct/24
cos x =c sin x =(√(1−c^2 )) ((√(1−c^2 )))^8 +c^8 = 2c^8 −4c^6 +6c^4 −4c^2 +1= =2(c^4 −c^2 +1+((√2)/2))(c^4 −c^2 +1−((√2)/2))= [c^4 −c^2 =−c^2 (1−c^2 )=−cos^2 x sin^2 x =((cos 4x −1)/8)] =2(((cos 4x)/8)+(7/8)+((√2)/2))(((cos 4x)/8)+(7/8)−((√2)/2))= =((cos^2 4x)/(32))+((7cos 4x)/(16))+((17)/(32))= =((1+cos 8x)/(64))+((7cos 4x)/(16))+((17)/(32))= =((cos 8x)/(64))+((7cos 4x)/(16))+((35)/(64))
cosx=csinx=1−c2(1−c2)8+c8=2c8−4c6+6c4−4c2+1==2(c4−c2+1+22)(c4−c2+1−22)=[c4−c2=−c2(1−c2)=−cos2xsin2x=cos4x−18]=2(cos4x8+78+22)(cos4x8+78−22)==cos24x32+7cos4x16+1732==1+cos8x64+7cos4x16+1732==cos8x64+7cos4x16+3564

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